0g of KCl will precipitate from each beaker.
Now, 100 mL of 4M HCl and 100 mL of 3.7M KCl are combined in the first beaker to create a total mole of Cl of (0.4 + 0.37) moles = 0.77 moles.
K+ stays at a total of 0.37 moles.
Total solution volume is equal to (100mL + 100mL)/1000mL, or 200mL, or 0.2L.
Total Cl per Liter = 0.77moles/0.2L = 3.85M Cl
total moles of K+ in a liter = 0.37 moles/0.2 liters = 1.85 million K+
Precipitation cannot occur unless Qsp is more than or equal to Ksp, i.e. Qsp ≥ Ksp
Qsp = [K][Cl] = [1.85]
No KCl will precipitate in the first beaker because [3.85] = 7.12 is less than 13.69 (Ksp).
Calculating the mass precipitated is not necessary because there is no precipitate.
and the solution is 0g.
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